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Summarize the pertinent information oblained by applying the graphing stralegy and sketch the graph of \( y=f(x) \). \( f(x)= \) In \( (x+3)-2 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) has a local minimum at \( x= \) (Type an exact answer. Use a comma to separate answers as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is concave upward on the subinterval(s) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The function \( f \) is never concave upward. Select the correct choice bolow and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is concave downwald on the subinterval(s) (Iype your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The function \( f \) is never concave downward.

Ask by Coles Riley. in the United States
Mar 11,2025

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**Summary:** - **Domain:** \( x > -3 \) - **Vertical Asymptote:** \( x = -3 \) - **Monotonicity:** Strictly increasing on \( (-3, \infty) \) - **Local Minimum:** None - **Concavity:** Concave downward on \( (-3, \infty) \) **Answer Choices:** A. The function \( f \) has no local minimum. B. The function \( f \) is concave downward on \( (-3, \infty) \).

Solution

**Domain and Asymptotes** - The function is given by \[ f(x)=\ln(x+3)-2. \] - Its domain is determined by the logarithm’s argument: \[ x+3>0\quad\Longrightarrow\quad x>-3, \] so the domain is \(( -3, \infty )\). - There is a vertical asymptote at \( x=-3 \) because as \( x\to-3^+ \), \[ \ln(x+3)\to-\infty. \] **Monotonicity and Local Extrema** - The first derivative is \[ f'(x)=\frac{1}{x+3}. \] - Since \( x+3>0 \) for all \( x>-3 \), we have \[ f'(x)>0, \] which implies that \( f \) is strictly increasing on its entire domain. - Because the function is strictly increasing and no critical points occur (i.e. \( f'(x)\neq 0 \) for any \( x \) in the domain), there is no local (or global) minimum _attained_ by \( f \) on \(( -3, \infty )\). **Concavity** - The second derivative is \[ f''(x)=-\frac{1}{(x+3)^2}. \] - Since \((x+3)^2>0\) for all \( x>-3 \), it follows that \[ f''(x)<0 \] for all \( x \) in the domain. - Therefore, the function is concave downward on the entire interval \(( -3, \infty )\) and never concave upward. **Graph Sketch Summary** - **Behavior near the vertical asymptote:** As \( x \to -3^+ \), \( f(x) \to -\infty \). - **Overall trend:** The function is strictly increasing. - **End behavior:** As \( x \to \infty \), \( f(x) \to \infty \). - **Concavity:** The graph is concave downward everywhere on its domain. **Answer Choices** 1. **Local Minimum:** The function \( f \) has no local minimum. 2. **Concavity Upward:** The function \( f \) is never concave upward. 3. **Concavity Downward:** The function \( f \) is concave downward on the subinterval \[ (-3, \infty). \]

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To understand the function \( f(x) = \ln(x+3) - 2 \), we first identify its domain, which is \( x > -3 \). The natural logarithm will tell us that this function has a vertical asymptote at \( x = -3 \) and will approach negative infinity as it nears this point. The graph will cross the horizontal line \( y = -2 \) when \( x = 0 \) because \( f(0) = \ln(3) - 2 \). Overall, scheming this out reveals the behavior and shifts due to the logarithm. For the local minimum: The function has a local minimum at \( x = 0 \) since \( f'(x) = \frac{1}{x+3} \) is positive for \( x > -3 \) and approaches zero as \( x \) approaches infinity. Regarding concavity: The function is concave down everywhere since \( f''(x) = -\frac{1}{(x+3)^2} < 0 \) for all valid \( x > -3 \). Therefore, we can say: - Local minimum at \( x=0 \) - Concave downward on the interval \( (-3, \infty) \)

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